Prove that limx→0[arcsin⁡xx]=1 and limx→0[arctan⁡xx]=0 where [] represents the greatest integer function.

For the second one, considerf(x)=arctan⁡xxAnd g(x)=arctan⁡x while h(x)=xLet's findlimx→0+arctan⁡xxOn applying L'Hospital rule we getlimx→0+arctan⁡xx=limx→0+11+x2→11+→1−On checking the limit of f(x) as x→0−, we similarly getlimx→0+arctan⁡xx→1−So it is pretty easy to notice that if we apply the greatest integer function to f(x) , then the value which reaches the floor function will be 1− which will be converted to absolute 0 by the floor function. Hencelimx→0[f(x)]=0Try the same with the first one

Expert Answer to "Prove that limx→0[arcsin⁡xx]=1 and limx→0[arctan⁡xx]=0 where [] represents..."

Marzadri9lyy

Answered question

2022-03-30

Prove that

lim_{x \to 0} [\frac{\arcsin x}{x}] = 1

and

lim_{x \to 0} [\frac{\arctan x}{x}] = 0

where [] represents the greatest integer function.

Answer & Explanation

Korbin Ochoa

Beginner2022-03-31Added 11 answers

For the second one, consider

f (x) = \frac{\arctan x}{x}

And

g (x) = \arctan x

while h(x)=x
Let's find

lim_{x \to 0^{+}} \frac{\arctan x}{x}

On applying L'Hospital rule we get

lim_{x \to 0^{+}} \frac{\arctan x}{x} = lim_{x \to 0^{+}} \frac{1}{1 + x^{2}} \to \frac{1}{1^{+}} \to 1^{-}

On checking the limit of f(x) as

x \to 0^{-}

, we similarly get

lim_{x \to 0^{+}} \frac{\arctan x}{x} \to 1^{-}

So it is pretty easy to notice that if we apply the greatest integer function to f(x) , then the value which reaches the floor function will be

1^{-}

which will be converted to absolute 0 by the floor function. Hence

lim_{x \to 0} [f (x)] = 0

Try the same with the first one

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